The question comes from the notion of a computable number. If we consider the set $$R=\{x\in\mathbb{R}_{\geqslant 0}: n\mapsto\lfloor nx\rfloor\text{ is recursive}\}$$ (with $n$ running over nonnegative integers, and the conventional definition of a recursive function), then $R\cup(-R)$ forms a field (correct me if I'm wrong). I'm wondering what exactly breaks down if we restrict ourselves to $n\mapsto\lfloor nx\rfloor$ being primitive recursive. If we call the resulting set $P$ (as in the title), is $P$ closed under addition (that is, does $x,y\in P\implies x+y\in P$)?
(A side note. The inclusion $P\subset R$ is proper; if $x=0.\overline{a_1a_2a_3\ldots}_2$ is the binary fraction representation of $x\in(0,1)$, then $a_n=\lfloor 2^n x\rfloor-2\lfloor 2^{n-1}x\rfloor$; thus, if $n\mapsto\lfloor nx\rfloor$ is [primitive] recursive, then so is $n\mapsto a_n$. And there are recursive functions $\mathbb{N}\to\{0,1\}$ which are not primitive recursive.)
I'm pretty sure the answer is negative, but I feel lost for seeing a proof. (Is there a way to attack it by relating $\lfloor nx\rfloor-n\lfloor x\rfloor$ to the indicator function of some non-recursive set?..) I'm a beginner in this field, but I'm aware of the basics. (It may well be a duplicate of another question; please indicate if it is.)